Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200000 + 7;
int n, sy, a, b, l[maxn], r[maxn], q;
double sum[maxn];
int find(double tar) {
  int L = 0, R = n + 1, mid;
  while (L + 1 < R) {
    mid = L + R >> 1;
    if (l[mid] < tar)
      L = mid;
    else
      R = mid;
  }
  return L;
}
inline double ip(double x1, double y1, double x2, double y2) {
  return x1 + (x1 - x2) * (-y1) / (y1 - y2);
}
double cal(int x, int y) {
  double p1 = ip(a, sy, x, y), p2 = ip(b, sy, x, y);
  int pp1 = find(p1), pp2 = find(p2);
  double ret = sum[pp2] - sum[pp1];
  if (p1 < r[pp1]) ret += r[pp1] - p1;
  if (p2 < r[pp2]) ret += p2 - r[pp2];
  return ret / (p2 - p1) * (b - a);
}
int main() {
  scanf("%d%d%d%d", &sy, &a, &b, &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &l[i], &r[i]);
  }
  for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + r[i] - l[i];
  cin >> q;
  for (int i = 1, x, y; i <= q; ++i) {
    scanf("%d%d", &x, &y);
    printf("%lf\n", cal(x, y));
  }
  return 0;
}